STEP Maths I, II, III 1994 Solutions

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Reply 40

That works :smile:

(Though for the first part, you can do it backwards.
$\frac{1-\cos{\alpha}}{\sin \alpha} = \frac{1-(1-2\sin^2 \frac{1}{2}\alpha)}{2 \sin{\frac{1}{2}\alpha}{\cos{\frac{1}{2}\alpha}}} = \frac{\sin \frac{1}{2}\alpha}{\cos{\frac{1}{2}\alpha}}$
Which is a little easier.)

Reply 41

DFranklin

Rabite

That works :smile:

Not only easier, but to be honest, I'd be a little antsy about all the square roots in the other method. You really need to prove you don't need to worry about all the implict $\pm$ signs.

Reply 42

khaixiang

nota bene

STEP III Q5

So we have $(1-0)^2f^{2k+3}(0)-(2k+1)^2f^{2k+1}(0)=0$ . Meaning: $f^{2(k+1)+1}(0)=(2k+1)^2f^{2k+1}$ Where k= 0, 1, 2...
So the coefficients of the terms are
$a_1=1 \newline a_3=\frac{1}{3!} \newline a_5=\frac{3^2}{5!}=\frac{3}{40} \newline a_7=\frac{5^2\times3^2}{7!}=\frac{5\times 3}{7\times6\times4\times2}$
etc. so the MacLaurin expansion is $arcsin(x)=x+\frac{1}{6}x^3+\frac{3}{40}x^5+\frac{5}{196}x^7+...$

To find the power series expansion for $g(x)=Ln(\sqrt{\frac{1+x}{1-x}})$ we can see it as $\frac{1}{2}[Ln(1+x)-Ln(1+(-x))]$ for which we can use the McLaurin series for Ln(1+x).
$Ln(1+x)= x - \frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^{n+1}\frac{x^n}{n}$
So: $\frac{1}{2}[(x - \frac{x^2}{2}+\frac{x^3}{3}+...+(-1)^{n+1}\frac{x^n}{n})- (-x - \frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^{n+1}\frac{(-x)^n}{n})]$
Even terms cancel and leaves $\frac{1}{2}\times 2 (x+\frac{x^3}{3}+\frac{x^5}{5}+...+\frac{x^{2n+1}}{2n+1})$
This means the the power series for g(x) is $\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$ and thus the coefficient is $\frac{1}{2n+1}$
Comparing coefficients of the series for f(x) and g(x) we can see that g(x)>f(x). For example the coefficients of $x^7$ which are $\frac{1}{7}>\frac{5\times 3}{7\times6\times4\times2}$ for g(x) and f(x) respectively.

Do I need a better justification for the coefficients of g(x) being larger? I find it a bit hard to express a general term for the power series of f(x). (well I can get it off wikipedia obviously but I have no idea how to get there and especially how to see it during a test).

As an alternative, earlier you've determined that

$f^{2(k+1)+1}(0)=(2k+1)^2f^{2k+1}$ Where k= 0, 1, 2...
or
$f^{2(k+1)+1}(0)=(2k+1)(2k+1)f^{2k+1}$ Where k= 0, 1, 2...

Then note that (by comparing the maclaurin series of g(x) that you've found with the $g(x)=g(0)+g'(0)x+\frac{g''(0)x^2}{2!}+\dots$ :

$g^{2(k+1)+1}(0)=(2k+1)(2k+2)g^{2k+1}$ Where k= 0, 1, 2...

Since g(0)=f(0), it soon follows that for k>1,

$\frac{g^{2k+1}(0)}{(2k+1)!}>\frac{f^{2k+1}(0)}{(2k+1)!}$

i.e the coefficients of maclaurin series of g(x) is greater than that of f(x) except for the first term.

And nota bene, you've done a great job for this question.

Reply 43

*bobo*

STEP III
(ds/dt)^2 + 2gy must always be constant as this equation forms the energy equation (incl. kinetic and GP) multiplied by 2 and divided by m, which is a constant. this equation must be a constant as no external forces are applied to the particle so energy is conserved.

(ds/dt)^2 + 2gk^(-1)(s/2)^2=c
so:
2(ds/dt)(d^2s/dt^2) + (sg/k)(ds/dt)=0
=> 2(d^2s/dt^2) + (sg/k)=0
=> (d^2/dt^2)= -(g/2k)s
so angular speed= (g/2k)^0.5
so period= 2pi(2k/g)^(0.5)
time taken to reach V is 0.25 of period= pi(k/2g)^(0.5)

Reply 44

nota bene

khaixiang

$g^{2(k+1)+1}(0)=(2k+1)(2k+2)g^{2k+1}$ Where k= 0, 1, 2...

|...|

And nota bene, you've done a great job for this question.

I knew it was something that I missed! Thanks, that seems to be the 'simple' way of doing it...
I quite liked this question, it doesn't really have anything complicated in it... :confused:

Although it turned out to be a bit long, but not too bad.

*bobo* are you going to add the finishing off on that question? (just looked and there is something with "describe the motion" of the m and $m\alpha$ thingies...). Good job either way :smile:

Reply 45

*bobo*

yes just had to go for my tea thoguh

Reply 46

*bobo*

if the particles are perfectly elastic then no energy should be lost in the collision, hence the total energy of the particles must be equal to 2ghk(1 + alpha^2). This was deduced by forming the energy equations of each particle and the constants.

Reply 47

*bobo*

STEP II
9) centre of mass from B is a distance (2a/3)
let A be the angle between OA and OB
bsinA/sin(90-0.5A)=2a
=>bcos0.5Asin0.5A=acos0.5A
=>sin0.5A=a/b

O must be directly above centre of mass so:
bsin(90-0.5A)/costheta + (2a/3)sintheta= bsin(90-0.5A+ theta)
=>(b^2- a^2)^0.5/costheta +(2a/3)sintheta= b(cos0.5Acostheta + sin0.5Asintheta)
=(b^2-a^2)^0.5costheta + asintheta
=> (a/3)sintheta= (b^2-a^2)^0.5sin^2theta/costheta
=>tan theta= a/(3(b^2-a^2)^0.5)

Reply 48

*bobo*

moments aboutB:
mgcos theta= T(b^2 -a^2)^0.5/b

sin theta= costheta a/ (3(b^2-a^2)^0.5
1-cos^2theta= (acos^2theta)^2/ (9(b^2-a^2)

=> 1- T^2(b^2-a^2)/(mgb)^2= a^2T^2/9(mgb)^2
=>T^2= 9(mbg)^2/(9b^2-8a^2)
T= 3mgb/(9b^2-8a^2)^0.5

Reply 49

*bobo*

STEP II
questions 9 and 10

STEPII q9and10.doc641.0KB

Reply 50

*bobo*

q 11

Doc4.doc520.2KB

Reply 51

*bobo*

11 cont. couldnt upload full attachment

Reply 52

and 3rd part to 11

Reply 53

STEP I Q 12

i) $\frac{2}{28}=\frac{1}{14}$

ii) $\frac{1}{28}\times1\times1+\frac{27}{28}\times \frac{1}{27}\times1+\frac{27}{28}\times\frac{26}{27}\times1=\frac{3}{28}$

iii) 1-(probability of failure) Probability of failure: $\frac{26}{28}\times\frac{25}{27}\frac{24}{26}= \frac{50}{63}$ So the answer is $\frac{13}{63}$

iv) $1\times \frac{1}{27}\frac{26}{26}+1\times \frac{26}{27} \times\frac{1}{26}=\frac{2}{27}$

v) P(Newnham+New Hall|Newnham)= $\frac{P(N\cap NH)}{P(N)}$ Then we know P(N) from ii) and $P(N\cap NH)=\frac{26}{28}\frac{2}{27}\frac{1}{26}+\frac{2}{28}\frac{26}{27}\frac{1}{26}+\frac{2}{28}\frac{1}{27}\frac{26}{26}=\frac{6}{27\times28}$
From that follows $\frac{P(N\cap NH)}{P(N)}=\frac{\frac{6}{27\times28}}{\frac{3}{28}}=\frac{2}{27}$

vi) Same as above, both probabilities double and hence cancel.

vii) $P(two|one)=\frac{P(one\cap \textbb{the other})}{P(one)}$ Now we already have the probabilities needed, from the answer in iii) and the $\frac{6}{27\times28}$ from v). So $\frac{\frac{6}{27\times28}}{\frac{13}{63}}=\frac{1}{26}$

Reply 54

ad absurdum

STEP II question 1

i)work mod 10. So the possible residues of n mod10 are 1,3,7,9. It suffices to show that there exists m mod10 such that n*m=1 (mod10) for each residue of n, so taking m=1,7,3,9 respectively gives the required result.

ii)If n ends in 1,3,7 or 9 we have shown we can multiply by some m to give a number ending in 1. So we only have to worry about n ending in 0,2,4,5,6,8,10. Now, clearly our number is divisible by 2 or 5, so suppose it is divisible by 2 and 5 a certain number of times each. All we have do is find an m containing the correct number of 2s or 5s such that the product n*m contains the same number of factors of 2 and 5. This means that n*m is divisble by 10 this number of times, and furthermore it is not divisible by 2 or 5 any further and hence the last digit is either 1,3,7 or 9. Applying the first part again gives the required result.

iii)Suppose n is a k-digit number as described. Using the number m=900..008000....00020..0010, where there are at least k+1 zeros between the adjacent 9,8,..,2,1 will work.

Reply 55

ad absurdum

STEP II question 6

Proof by induction. For n=1:
$RHS =\frac{1}{2}\cot\frac{\theta}{2} - \cot\theta$
$\frac{1}{2} \cot\frac{\theta}{2} - \frac{1-\tan^2\frac{\theta}{2}}{\tan\frac{theta}{2}}$
$\frac{1}{2} \tan\frac{theta}{2} = LHS$ , as required

Assume the given formula holds. So, for n+1,
$\frac{1}{2} \tan\frac{\theta}{2} + \frac{1}{2^2} \tan\frac{\theta}{2^2} + ... + \frac{1}{2^{n+1}} \tan\frac{\theta}{2^{n+1}}}$
$= \frac{1}{2^n}}\cot\frac{\theta}{2^n} - \cot\theta + \frac{1}{2^{n+1}}} \tan\frac{\theta}{2^{n+1}}$
$= \frac{1}{2^n}(\frac{1-tan^2\frac{\theta}{2^{n+1}}}{2tan\frac{\theta}{2^{n+1}}} + \frac{1}{2}\tan\frac{\theta}{2^{n+1}}}) - \cot\theta$
$=\frac{1}{2^{n+1}}\cot\frac{\theta}{2^{n+1}}} - \cot\theta$ , completing the proof by induction.

For the last part, note that as n-> infinity, the angle we are taking the cot of gets very small, and as sin(theta)-->theta and cos(theta)-->1, we can write the RHS for large n as:
$\frac{1}{2^n} \frac{2^n}{\theta} - \cot\theta = \frac{1}{\theta} - \cot\theta$ , as required.

Reply 56

zrancis

STEP I - Question 11

i)

To ensure the ball hits the ground, it's mass must be greater than the parallel component of the wagon:

$M>mgsin\theta$

By considering both bodies separately, there will be a tension pulling the wagon up the slope and a tension resisting the ball from falling, by applying N2L, firstly to the wagon:

$T-mgsin\theta=ma_c$

and to the ball:

$Mg-T=Ma_c$

Adding these gives:

$Mg-mgsin\theta=a_c(M+m)$

So:

$a_c=\frac{g(M-msin\theta)}{M+m}$

By taking the length of the slope to be: $s=dsin\theta$
And applying the fact the system starts from rest,
Suvat can be applied:

$v^2=u^2+2a_cs$

$v^2=0+2\frac{g(M-msin\theta)}{M+m}dsin\theta$

$v^2=\frac{2g(M-msin\theta)dsin\theta}{M+m}$

As required.

ii)

If the wagon is not to collide with the pulley, then as the ball is on the point of hitting the floor, it must have a velocity that is less than or equal to 0:

$v^2\le0$

hence:

$\frac{2g(M-msin\theta)dsin\theta}{M+m}\le0$

$sin\theta(M-msin\theta)\le0$

So:

Provided $sin\theta\ge0$

$M-msin\theta\le0$

QED (I hope)

I'm just in two minds about my reasoning in the last part, if somebody could take a quick look at it that'd be great.

Reply 57

Kolya

I would be interested if anybody could show how to do STEP I Q8. I get the impression that you must use the rather-too-coincidental limits to evaluate the integrals, but I cannot see how. Can anybody help?

Reply 58

insparato

Hmm i cant see where to go on this the change in variable doesnt really help at all.

$\displaystyle \int_0^{\frac{\pi}{4}} ln (1+tan\theta) \hspace5 d\theta$

$\theta = \frac{\pi}{4} - \phi$

$\frac{d\theta}{d\phi} = -1$

$d\theta = -d\phi$

$\displaystyle -\int_{\frac{\pi}{4}}^0 ln (1 + tan (\frac{\pi}{4} - \phi) ) \hspace5 d\phi$

$\displaystyle -\int_{\frac{\pi}{4}}^0 ln \left(1 + \frac{tan \frac{\pi}{4} - tan\phi}{1 + tan\frac{\pi}{4}tan\phi}\right)$

$\displaystyle -\int_{\frac{\pi}{4}}^0 ln \left(\frac{1 + tan\phi + 1 - tan\phi}{1 + tan\phi}\right)$

$\displaystyle -\int_{\frac{\pi}{4}}^0 ln \left(\frac{2}{1+tan\phi}\right)$

$\displaystyle \int_{\frac{\pi}{4}}^0 ln \left(\frac{1+tan\phi}{2}\right)$

Reply 59

Speleo

I = INT (0 to pi/4) ln(1+tanx) dx
let y = pi/4 - x
dy/dx = -1

-> - INT (pi/4 to 0) ln(1+tan(pi/4 - y)) dy
= INT (0 to pi/4) ln(1+ tan(pi/4 - y)) dy
= INT ln(1 + [1-tany/1+tany]) dy
= INT ln(2/1+tany) dy
I = INT ln(2/1+tanx) dx
Add original integral:
2I = INT (0 to pi/4) ln(1+tanx) + ln(2/1+tanx) dx
= INT ln2 dx
I = (1/2) INT (0 to pi/4) ln2 dx
= (1/2)(pi/4)ln2

= pi.ln2/8

Tell me if you want me to do the other parts too.

EDIT: glancing at the other parts, the second one looks like x = tanu transforms it to the original integral or very close, the last one is a y = pi/2 - x substitution, noting that sin and cos can be swapped over with the given limits, and applying a similar trick to get a simple integral (adding the original integral).

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STEP Maths I, II, III 1994 Solutions

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